How many words can be formed by taking 4 letters at a time from the letters of the word 'MORADABAD'?

Solution:

There are 9 letters in the word MORADABAD, namely AAA, DD, M, R, B and O.

The four-letter word may consists of

(i) 3 alike letters and 1 distinct letter

(ii) 2 alike letters of one kind and 2 alike letters of the other kind

(iii) 2 alike letters and 2 distinct letters

(iv) all different letters

(i) 3 alike letters and 1 distinct letter:

There is one set of three alike letters, AAA, which can be selected in one way.

Out of the 5 different letters $D, M, R, B$ and $O$, one can be selected in $^{5} C_{1}$ ways.

These four letters can be arranged in $\frac{4 !}{3 ! 1 !}$ ways.

 

$\therefore$ Total number of ways $={ }^{5} C_{1} \times \frac{4 !}{3 ! 1 !}=20$

(ii) There are two sets of two alike letters, which can be selected in 2C2 ways.

Now, the letters of each group can be arranged in $\frac{4 !}{2 ! 2 !}$ ways.

 

$\therefore$ Total number of ways $={ }^{2} C_{2} \times \frac{4 !}{2 ! 2 !}=6$

(iii) There is only one set of two alike letters, which can be selected in 2C1 ways.

Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.

Thus, 2 alike letters and 2 different letters can be selected in ${ }^{2} C_{1} \times{ }^{5} C_{2}=20$ ways.

Now, the letters of each group can be arranged in $\frac{4 !}{2 !}$ ways.

 

$\therefore$ Total number of ways $=20 \times \frac{4 !}{2 !}=240$

(iv) There are 6 different letters A, D, M,B, O and R.

So, the number of ways of selecting 4 letters is 6C4, i.e. 15, and these letters can be arranged in 4! ways.

$\therefore$ Total number of ways $=15 \times 4 !=360$

 

$\therefore$ Total number of ways $=20+6+240+360=626$